Search found 1015 matches
- Sat Dec 17, 2011 9:16 am
- Forum: Junior Level
- Topic: Tom's Mysterious Quadrilateral [Self-Made]
- Replies: 6
- Views: 4786
Tom's Mysterious Quadrilateral [Self-Made]
One day Tom decided to have some fun with geometry. So he drew a quadrilateral $ABCD$. Suddenly he noticed that his quadrilateral has a very nice property: $AB\cdot CD=BC\cdot AD$,$AB\cdot BC=CD\cdot AD$.It made him curious and he marked mid-points of side $AB,BC,CD,AD$ as $E,F,G,H$,respectively. \T...
SUM [OWN]
Let $t_0=1$ and for $n\ge1$,$t_n=\sum^{n-1}_{k=0} {}^{n-1}P_{n-1-k}2^{n-1-k}t_k$.Prove that
\[\sum^{\infty}_{j=0}$ $\left (\frac{t_j}{t_{j+1}}\right )^2=\frac {\pi^2}{8}$.
\[\sum^{\infty}_{j=0}$ $\left (\frac{t_j}{t_{j+1}}\right )^2=\frac {\pi^2}{8}$.
- Fri Dec 16, 2011 10:43 am
- Forum: Secondary Level
- Topic: Factorizing
- Replies: 5
- Views: 4205
Re: Factorizing
Dirichlet's theorem(the one I used here) states that if $p$ is an odd prime, $a$ is a co-prime of it and let $b=(p-1)/2$, then $a^b\equiv 1(mod p)$ happens if $x^2\equiv a(mod p)$ has solution.If this modular equation doesn't have any solution, then $a^b\equiv -1(mod p)$.Proof of this theorem can be...
- Thu Dec 15, 2011 3:53 pm
- Forum: Secondary Level
- Topic: Factorizing
- Replies: 5
- Views: 4205
Re: Factorizing
I first tried with Fermat's little theorem but did not succeed. I needed to use a theorem of Dirichlet. Notice that \[8=(17-1)/2\]Since there is no solution of the modular equation $x^2\equiv 3 (\bmod17),$ we can say from Dirichlet's theorem that $ 3^8\equiv -1 (\bmod17).$ But $x^2\equiv 2 (mod 17)$...
- Sun Dec 04, 2011 11:10 am
- Forum: Number Theory
- Topic: Find non-negative integers(mine)
- Replies: 9
- Views: 5868
Re: Find non-negative integers(mine)
Let $7^a+11^b=x^2$ It is very easy to see a is even and b is odd.(Use mod 3 and mod 4) Now let $a=2k$ Then,$11^b=(x+7^k)(x-7^k)$ Now look that $x-7^k$ is either 1 or a power of 11. If it is 1, then, $7^a+11^b=7^a+2 \cdot 7^k+1$ $11^b-1=2 \cdot 7^k$ LHS is containing 5 as factor,but RHS not.It is not...